10 mL of gaseous hydrocarbon on combustion gives 40 mL of $$CO_2(g)$$ and 50 mL of water vapour. Total number of carbon and hydrogen atoms in the hydrocarbon is ______.

10 mL of gaseous hydrocarbon on combustion gives 40 mL of $$CO_2$$ and 50 mL of water vapour (all at same T and P). We need the total number of carbon and hydrogen atoms in the hydrocarbon.

For a hydrocarbon $$C_xH_y$$:

$$C_xH_y + \left(x + \frac{y}{4}\right)O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O$$

At the same temperature and pressure, equal volumes of gases contain equal numbers of molecules. So the volume ratios equal the mole ratios.

From 10 mL of $$C_xH_y$$:

- $$CO_2$$ produced = 40 mL, so $$10x = 40$$, giving $$x = 4$$

- $$H_2O$$ produced = 50 mL, so $$10 \times \frac{y}{2} = 50$$, giving $$y = 10$$

The hydrocarbon is $$C_4H_{10}$$ (butane).

Total number of carbon and hydrogen atoms = $$x + y = 4 + 10 = 14$$.

The answer is 14.